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Math Help - Exponential radioactive decay while original amount increases.

  1. #1
    Newbie wolf's Avatar
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    Exponential radioactive decay while original amount increases.

    This problem is much trickier than I thought.
    I can solve the "easy" half-life problems, such as beginning amount, ending amount and so on. This one's tougher.

    Let's suppose you have a substance "X" that has a half life of 30 days. You start out at day zero with 0 grams of "X". For each day, 1 gram of "X" is added to a beaker. (This is a continual process whereby, at the end of each hour, the amount of "X" is increased by 1/24 of a gram, or for each second the amount is increased by 1/86,400 gram. The increases are not done by "chunks".)
    Okay so each day, you have an additional gram of "X" and after 5 days (for example) you have 5 grams of "X". But this isn't exactly accurate because some of that 5 grams has already started to decay with a half-life of 30 days.

    So, what is the formula for determining how much "X" is present at any particular time?
    And since this is my first posting - hello everyone.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by wolf View Post
    This problem is much trickier than I thought.
    I can solve the "easy" half-life problems, such as beginning amount, ending amount and so on. This one's tougher.

    Let's suppose you have a substance "X" that has a half life of 30 days. You start out at day zero with 0 grams of "X". For each day, 1 gram of "X" is added to a beaker. (This is a continual process whereby, at the end of each hour, the amount of "X" is increased by 1/24 of a gram, or for each second the amount is increased by 1/86,400 gram. The increases are not done by "chunks".)
    Okay so each day, you have an additional gram of "X" and after 5 days (for example) you have 5 grams of "X". But this isn't exactly accurate because some of that 5 grams has already started to decay with a half-life of 30 days.

    So, what is the formula for determining how much "X" is present at any particular time?
    And since this is my first posting - hello everyone.
    Make a function such as

    f(t)=\text{initial amount plus amount added}-\text{amount accumitavely decayed}
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  3. #3
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    Quote Originally Posted by wolf View Post
    This problem is much trickier than I thought.
    I can solve the "easy" half-life problems, such as beginning amount, ending amount and so on. This one's tougher.

    Let's suppose you have a substance "X" that has a half life of 30 days. You start out at day zero with 0 grams of "X". For each day, 1 gram of "X" is added to a beaker. (This is a continual process whereby, at the end of each hour, the amount of "X" is increased by 1/24 of a gram, or for each second the amount is increased by 1/86,400 gram. The increases are not done by "chunks".)
    Okay so each day, you have an additional gram of "X" and after 5 days (for example) you have 5 grams of "X". But this isn't exactly accurate because some of that 5 grams has already started to decay with a half-life of 30 days.

    So, what is the formula for determining how much "X" is present at any particular time?
    And since this is my first posting - hello everyone.

    This would be a first order linear differential equation

    Since the substance is not added in chunks lets suppose that is it added continously. Then the rate in would be

    r_i=1 \frac{gram}{day}

    The rate out would be

    r_0=kA where A is the number of grams of the substance at any time (t) and k=\frac{\ln(2)}{30}

    \frac{dA}{dt}=1-\frac{\ln(2)}{30}A

    \frac{dA}{dt}+\frac{\ln(2)}{30}A=1

    Our integrating factor is e^{\frac{\ln(2)}{30}t}

    \frac{d}{dt}\left[e^{\frac{\ln(2)}{30}t}A \right]=e^{\frac{\ln(2)}{30}t}

    Integrating both sides and solving for A we get

    A(t)=\frac{30}{\ln(2)}+Ce^{-\frac{\ln(2)}{30}t}

    Using your inital condition that A(0)=0 we get C=-\frac{30}{\ln(2)}

    A(t)=\frac{30}{\ln(2)}-\frac{30}{\ln(2)}e^{-\frac{\ln(2)}{30}t}

    t \to \infty \\\ A(t) \to \frac{30}{\ln(2)} \approx 43.281
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  4. #4
    Newbie wolf's Avatar
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    Wow, those replies came pretty fast.

    Let's see if I can work out the calculations of the formula for 1 gram per day, for 5 days with a 30 day half life.

    30/ln(2) -30/ln(2) * e^-(ln(2)/30)*5 =

    43.28085 - 43.28085*e^-(ln(2)/30)*5 =

    43.28085 - 43.28085*2.718281828^-((.02310491)*5) =

    43.28085 - 43.28085*2.718281828^-(.11552453) =

    43.28085 - 43.28085*.8908987 =

    43.28085 - 38.558853 =

    4.721997 grams of "X" after 5 days

    That appears to be correct. Did I interpret the formula correctly?



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    Behold, the power of SARDINES!
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    Thumbs up

    Quote Originally Posted by wolf View Post
    Wow, those replies came pretty fast.

    Let's see if I can work out the calculations of the formula for 1 gram per day, for 5 days with a 30 day half life.

    30/ln(2) -30/ln(2) * e^-(ln(2)/30)*5 =

    43.28085 - 43.28085*e^-(ln(2)/30)*5 =

    43.28085 - 43.28085*2.718281828^-((.02310491)*5) =

    43.28085 - 43.28085*2.718281828^-(.11552453) =

    43.28085 - 43.28085*.8908987 =

    43.28085 - 38.558853 =

    4.721997 grams of "X" after 5 days

    That appears to be correct. Did I interpret the formula correctly?
    Yes that is correct.
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  6. #6
    Newbie wolf's Avatar
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    Okay, and thanks to all for helping me.
    This message board is incredible.
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